1.1 Textbook Learning Objectives

• Recognize central limit theorem problems.
• Classify continuous word problems by their distributions.
• Apply and interpret the central limit theorem for means.
• Apply and interpret the central limit theorem for sums.

1.2 Instructor Learning Objectives

• Recognize the prevalence and importance of the mean
• Understand how the central limit theorem contributes to the accuracy of statistics in representing/estimating parameters.

2.1 Introduction

• Discuss: Prior to this next part, try writing notation to represent a continuous variable

• The central limit theorem around means can be represented in a similar notation to distribution like we used previously in Modules 4 and 5
  • This is because the central limit theorem hinges around a distribution of sample means
• The formula to represent the central limit theorem of means is: \(\bar{X} \sim N(\mu_X, \frac{\sigma_X}{\sqrt{n}})\)
• Where:
  • \(\bar{X}\) is a random continuous variable consisting of many sample means of variable \(X\)
  • \(\mu_X\) is the mean of random continuous variable \(X\)
  • \(\sigma_X\) is the standard deviation of random continuous variable \(X\)
  • \(n\) is the size of the individual samples taken to from the sampling distribution (this is consistent across all theoretical samples)

• Important: All of the above Xs are uppercase! This is because we are using it to describe a variable, not the possible values that the variable could take, which would be lowercase x.

• Discuss: Try interpreting the above notation like you would any other normal distribution, like the one you wrote in the last 'Discuss' question. Effectively, try to write out what each part of the notation means.

2.2 Sampling Distributions

• When we represent many sample means as a distribution, we call this the sampling distribution of means, which is really what is shown in the prior notation
  • A high standard deviation in this sampling distribution would suggest that the individual sample means are more spread out, i.e., each sample is more different from one another
  • Thus, a small standard deviation in the sampling distribution is ideal, as it indicates that each individual sample is relative close in its mean estimate
• The CLT holds that the larger a sample size, the more its sampling distribution will resemble a normal distribution
  • This hinges on \(n\) being large enough
  • Consider the formula again: \(\bar{X} \sim N(\mu_X, \frac{\sigma_X}{\sqrt{n}})\)
  • \(n\) being in the denominator means that a larger \(n\) will result in a smaller overall standard deviation for the distribution
  • This \(\frac{\sigma_X}{\sqrt{n}}\) is referred to as the standard error of the mean

• Important: You can think of the standard error as being the standard deviation of the sampling distribution, or how far away, on average sample means are away from the mean of sample means.

• Question: Earlier, what was the rule-of-thumb number for what is 'large enough' sample?
• A) 20
• B) 10
• C) 100
• D) 30

2.3 Z-scores for Sampling Distributions

• We can conceptualize the mean of a single sample as a z-score for how far away it is from the theoretical mean of the sampling distribution with:

\[ z = \frac{\bar{x} - \mu_X}{(\frac{\sigma_X}{\sqrt{n}})} \]

• Where:
  • \(\bar{x}\) is the mean for a single sample
  • \(\mu_X\) is the mean of both \(X\) and \(\bar{X}\)
  • \(\sigma\) same as before

2.4 Connection to Finding Probability in the Normal Distribution

• Because the sampling distribution is theorized as a normal distribution, we can estimate the probability that a certain event regarding an individual sample mean occurs
  • In order to do so, we need some information about the believed mean (\(\mu_X\)), standard deviation (or standard error; \(\sigma_X\)), and sample size (\(n\)) of the particular sampling distribution

2.5 Educational Example

• A standardized elementary reading test for children produces scores that range from 0 to 500, which represent student ability in reading for that grade level
  • Treat \(R\) as representing the continuous random variable of score on the reading test
• Based on the test manufacturers specifications, this test has a mean sampling distribution mean of 250, and standard deviation of 50. In my scenario, let’s assume I take samples of \(n = 40\).
  • My sampling distribution could be represented as:

\[ \bar{R} \sim N(250, \frac{50}{\sqrt{40}}) \]

• Where:
  • \(\bar{R}\) is the mean of sample means of the reading scores
• The z-score formula then could be shown as:

\[ z = \frac{\bar{r} - 250}{(\frac{50}{\sqrt{40}})} \]

• Where:
  • \(\bar{r}\) is a single sample mean

3.1 Brief Overview

• Important: I am going to focus less on this one, only because it ends up following a very similar idea and logic to the mean CLT.

• The central limit theorem for sums functions on a very similar logic to the CLT for means, just with a different formula to represent the distribution:

\[ \sum{X} ~ N((n)(\mu_X), (\sqrt{n})(\sigma_X)) \]

• Where:
  • \(\sigma{X}\) is the continuous variable of sums of \(X\)
  • \(\mu_X\) is the mean of \(X\)
  • \(\sigma_X\) is the standard deviation of \(X\)
• The z-score formula:

\[ z = \frac{\sum{x}-(n)(\mu_X)}{(\sqrt(n)(\sigma_X))} \]

• The end idea of this version has a similar gist: larger \(n\) narrows the standard error of the distribution of sums.

4.1 Introduction

• In practice, the central limit theorem is not often directly used, but is often appealed to as part of statistical inference
  • Most classic, parametric tests rely upon the implications of the central limit theorem
• The central limit theorem is best applied to aggregate values, such as mean and sums, not individual points
  • E.g., I don’t need to use the CLT to make sense of a single test score in a sample I take

• Important: Much like the normal distribution in general, CLT is sometimes misused as providing an easy appeal, but it's not perfect!

4.2 Connection to the Law of Large Numbers

• If the central limit theorem implies that the center of the distribution of sample means is the population mean parameter, then a larger \(n\) allows our \(\bar{x}\) to better estimate the \(\mu\).

• Important: Remember that this is assuming that our sample is representative of the population of interest!

• Discuss: What are some example of randomness-based sampling techniques that result in representative samples?

5.1 Recap

• The CLT serves as a critical concept building on probability and continuous, normal distributions

• The CLT underlies why larger samples result in more accurate estimates of our population parameters in our sample statistics

• The CLT implications and characteristics are useful in making sense of results from statistical hypothesis testing (more on this later!)

5.2 Lecture Check-in

• Make sure to complete any lecture check-in tasks associated with this lecture!

Module 7 Lecture - Factorial/Two-Way ANOVA || Analysis of Variance